Verified Solution • 2025–26 Session moderate

QJEEM19Ph10dc: JEE Main 2019 Physics - Class 11 JEEM19Ph10dc Solution: The energy required to take a satellite to a height 'h' above…

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 ×\times 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is

Multiple Choice Options

  • A 1.6 ×\times 103 km
  • B 3.2 ×\times 103 km
  • C 6.4 ×\times 103 km
  • D 1.28 ×\times 104 km

Step-by-Step Verified Solution

Energy required to move a satellite from earth surface to height h is,

E1 = Uh −- Usurface

= −- GMmR+h−(−GMmR){{GMm} \over {R + h}} - \left( { - {{GMm} \over R}} \right)

= GMm (1R−1R+h)\left( {{1 \over R} - {1 \over {R + h}}} \right)

= GMmR(R+h)×h{{GMm} \over {R(R + h)}} \times h

We know, for sattelite at height h.

Centrifigual force = Gravitational force

⇒\Rightarrow    mv2R+h=GMm(R+h)2{{m{v^2}} \over {R + h}} = {{GMm} \over {{{\left( {R + h} \right)}^2}}}

⇒\Rightarrow    mvmv 2 = GMmR+h{{GMm} \over {R + h}}

∴\therefore    12mv2{1 \over 2}m{v^2} = GMm2(R+h){{GMm} \over {2\left( {R + h} \right)}}

∴\therefore   Kinetic energy (E2) = GMm2(R+h){{GMm} \over {2(R + h)}}

Given that,

E1 = E2

∴\therefore    GMmR(R+h)×h=GMm2(R+h){{GMm} \over {R(R + h)}} \times h = {{GMm} \over {2\left( {R + h} \right)}}

⇒\Rightarrow    hR=12{h \over R} = {1 \over 2}

⇒\Rightarrow   h = R2{R \over 2}

∴\therefore   h = 6.4×1032{{6.4 \times {{10}^3}} \over 2}

= 3.2 ×\times 103 km

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